Calculating the Probability of Simple Events. In this video I show the basic idea and a few simple examples of calculating the probability of simple events!
For more free math videos, visit http://JustMathTutoring.com
Duration : 0:9:25
[youtube BAjOEsU_mpE]
February 5th, 2010 at 2:15 pm
i always get heads …
i always get heads
February 5th, 2010 at 2:15 pm
Bob arrives to work …
Bob arrives to work on time with probability 0.70, while Caroline arrives on times with probability 0.80. The probability they both arrive on time is 0.60.
1)What is the probability that at least one of them arrives on time?
2)What is the probability that exactly one will arrive at work on time?
3)What is the probability that neither one of them will arrive to work on time?
February 5th, 2010 at 2:15 pm
??????????????????? …
?????????????????????/
February 5th, 2010 at 2:15 pm
YOU ARE LEFT-HANDED
YOU ARE LEFT-HANDED
February 5th, 2010 at 2:15 pm
my brain hurts
my brain hurts
February 5th, 2010 at 2:15 pm
BUT THAT’S NOT THE …
BUT THAT’S NOT THE QUESTION! The question is the probability of flipping a coin TWICE and it being heads BOTH times. If the question was, what is he probablility of each flip..yeah 1/2. But that’s not the question is it? Two head in a row from the beginning. 1/4. God, this is getting even more frustrating.
February 5th, 2010 at 2:15 pm
Both of you are …
Both of you are right, but it depends on what event you look at: If you look at the series of flipping the coin twice, then the probability of that happening with the same result is 1/4. If you look at the probability of flipping the coin once, then it is 50/50 or 1/2.
February 5th, 2010 at 2:15 pm
or you would do the …
or you would do the inverse of 2 to the power of n, with n being the times you flip the coin.
February 5th, 2010 at 2:15 pm
question! could we …
question! could we potentially know the result of the coin toss definitively, if we know all the variables and constants influencing how it lands? the downwards pull of gravity… the weight of the coin… the force and direction in which you flip the coin (which is really what changes)… etc?
the side the coin lands on is more a a question of the physics involved, isn’t it? only we can’t predict the variables…
February 5th, 2010 at 2:15 pm
Help!! I’m a …
Help!! I’m a bartender in a local pub to work my way through school and (don’t ask how it got brought up) but I’ve been arguing with a patron for two solid weeks that the probability of flipping a coin twice and getting heads both times is 1/4. Easy enough right? Not for him apparantly. It has annoyed me to no end. His basic argument is this. After flipping the coin once if it ends tails the condition is not met. thus (and I know it’s insanely wrong) your odds are 50/50…help me!!!
February 5th, 2010 at 2:15 pm
good job
good job
February 5th, 2010 at 2:15 pm
I figured out the …
I figured out the coin one relatively fast, rather than doing all the math.
Chance to land a head is 50%, so with that in mind…
Imagine you flip the coins and land on heads every time, and when you do, you drink half of a full can of pop. You drink half (50%) if you get heads. Another half (25%) if you get heads again. And another half (12.5%).
12.5(8) = 100
February 5th, 2010 at 2:15 pm
yeah sry i was …
yeah sry i was thinking about the probability of getting 5 heads without the trick coin sry. Thank you!!!
February 5th, 2010 at 2:15 pm
Your method and …
Your method and reasoning is correct, turkeyman64, and my answer in exact form is consistent with your answer in exact form.
By independence and the addition theorem of mutually exclusive events, I evaluate ((9/10)(1/2)+(1/10)(1))^5 = ((9/20)+(1/10))^5 = (11/20)^5. This gives about 5%, though, not 3%, so your rounded answer is slightly off.
February 5th, 2010 at 2:15 pm
My logic is that …
My logic is that the inside of the outer parenthesis is what it is because there is a 9/10 chance that there will be a 50% chance of getting a head or (”or” my stats teach told me means add) there is a 1/10 chance that there will be a 100% chance of getting a head. So now that we have the probability of getting heads on one flip we need that 5 more times, so we need to flip a head and a head and a head and a head and a head (”and” my stats teacher tells me means multiply) there for i get ^5.
February 5th, 2010 at 2:15 pm
I have a …
I have a conditional probability problem that another probability youtuber has done that i am having trouble with how he got the answer. He asks, “if you have a bag filled with 9 regular coins and 1 double sided head coin, what is the probability of getting all heads for 5 flips. I thought the equation would end up being ((9/10)(.5)+(1/10)(1))^5. My next comment will explain my logic. his answer is somewhere around 70% but mine is near 3%, tell me if and if so why im wrong.
February 5th, 2010 at 2:15 pm
thank you patrick. …
thank you patrick. there’s a chance i can pass my exam i i keep on watching your videos…
February 5th, 2010 at 2:15 pm
Thank you so much …
it clarified some things that I was questioning about~
Thank you so much for making this video
February 5th, 2010 at 2:15 pm
cool vids man
cool vids man
February 5th, 2010 at 2:15 pm
runescape is good, …
runescape is good, im called Nick00179…Great guide
February 5th, 2010 at 2:15 pm
You have 3 doors, …
You have 3 doors,
you need to pick 2 out of 3 doors without hitting a X that is mark behind the door to win $10,000,you pick the first door and there is no X, now there are 2 doors left, one with a X and one without a X, what is your probability of hitting a door without a X?
February 5th, 2010 at 2:15 pm
Taking the …
Taking the leap-year in consideration.
A year averages 365.25 days – - (365+365+365+366)/4
February can have 28.25 – - (28+28+28+29)/4 days.
That’s 113/1461, or 7.73443%
0.0631% difference if you don’t take the leap year in consideration.
February 5th, 2010 at 2:15 pm
Great JOB………. …
Great JOB………….I PAT ur BACK……….PLZZZ PUT ON MORE PROBLEM AND SOLUTION VIDS OF PROBABILITY.THNX ALOT
February 5th, 2010 at 2:15 pm
daz actcully alsome …
daz actcully alsome nojoke thanks
February 5th, 2010 at 2:15 pm
this coming from a …
this coming from a person who has ‘runescape’ videos on their channel, loooooooool : )